Question

Pleaseeee help meeeeeWhat is ∑n=125(3n−2) equal to?

Answer 1

Answer:

925 is your answer!

Step-by-step explanation:

Hope this helps!

Answer 2

Answer:

$\sum_{n=1}^{25}(3n-2)=925]$

Step-by-step explanation:

The given series is $\sum_{n=1}^{25}(3n-2)$

The first term of this series is

$a_1=3(1)-2$

$a_1=3-2$

$a_1=1$

The second term is

$a_2=3(2)-2$

$a_2=6-2$

$a_2=4$

The common difference is

$d=4-1=3$

The sum of the first n-terms is given by;

$S_n=\frac{n}{2}[2a_1+d(n-1)]$

The sum of the first 25 terms of the series is

$S_{25}=\frac{25}{2}[2(1)+3(25-1)]$

$S_{25}=\frac{25}{2}[2+3(24)]$

$S_{25}=\frac{25}{2}(74)]$

$S_{25}=925]$